3 Answers
3

It is easy to see that at least one of $-1,2,-2$ is a square in that field: the set of primes where neither $-1$ nor $2$ is a quadratic residue is contained in the set of primes where $-2$ is a quadratic residue.

I was about to post the same (using 2,3, and 6, but, same thing)! Thanks.
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James CranchJan 28 '12 at 10:48

I believe one can prove that in that field every element is the sum of two squares. Also, its algebraic closure is isomorphic to the complex numbers (this is just a matter of cardinalities).
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Ramiro de la VegaJan 28 '12 at 12:37

A (non principal) ultraproduct of finite fields of unbounded cardinalities is
a pseudo-finite field :

It is perfect (if its characteristic is a prime number $p$, then any element
is a $p$th power) ;

For any positive integer $n$, it has exactly one extension of degree $n$
(equivalently, its Galois group is the profinite completion of the
integers $\mathbf Z$);

It is pseudo-algebraically closed (any geometrically irreducible algebraic
variety has a rational point).

Indeed, these are first-order properties, hence are inherited by ultraproducts :
the first two ones hold for finite fields, while, it follows from Lang-Weil estimates that the last one is satisfied if the cardinality of the field is large enough, depending on the degrees of the equations.

Observe that the field $\mathbf Q$ of rational numbers does not satisfy the last two properties.

In fact, it is a theorem of Ax (1968, The elementary theory of finite fields, Ann. Math. 88 (1968) 239-271) that conversely, pseudo-finite fields are elementary equivalent to ultraproduct of finite fields.

Regarding your edit, of course any ultrapower $\Pi\mathbb{Q}/U$ of $\mathbb{Q}$ itself is a nonstandard model of the theory of $\mathbb{Q}$ (in whatever language you choose). So this version of $\mathbb{Q}^\ast$ is an ultraroduct. Indeed, ultraproducts are one of the principal methods of constructing nonstandard models.